integral from 1 to 2 of e^x+1/x

Lösung

\int_{1}^{2} e^{x} + \frac{1}{x} d x

e^{2} - e + ln (2)

Dezimale

5.36392 \dots

Schritte zur Lösung

\int_{1}^{2} e^{x} + \frac{1}{x} d x

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{1}^{2} e^{x} d x + \int_{1}^{2} \frac{1}{x} d x

\int_{1}^{2} e^{x} d x = e^{2} - e

\int_{1}^{2} \frac{1}{x} d x = ln (2)

= e^{2} - e + ln (2)

\int_{0}^{1} e^{2 x - 1} d x

\int_{0}^{u} \frac{1}{x} d x

\int_{- 4}^{4} - 5 d x

\int_{2}^{4} 4 x^{2} - 5 x + 8 d x

\int_{0}^{3} \frac{x e ^{x}}{( x + 1 ) ^{2}} d x